Solve the given system of equations.
\[
\begin{array}{lr}
x+y+3 z= & 6 \\
x+y+7 z= & 18 \\
x+9 y+8 z= & -19
\end{array}
\]

Select the correct choice below and fill in any answer boxes within your choice.
A. There is one solution. The solution set is , $\square, \square)\}$. (Simplify your answers.)
B. There are infinitely many solutions.
C. There is no solution.

Step 1 ：Subtract equation 1 from equation 2, we get: \(4z = 12\)

Step 2 ：Solve for z: \(z = \frac{12}{4} = 3\)

Step 3 ：Substitute \(z = 3\) into equation 1 and equation 2, we get: \(x + y = 6 - 3*3 = -3\) and \(x + y = 18 - 3*7 = -3\)

Step 4 ：These two equations are the same, which means they are dependent.

Step 5 ：Substitute \(z = 3\) into equation 3, we get: \(x + 9y = -19 - 3*8 = -43\)

Step 6 ：Subtract equation 4 from equation 6, we get: \(8y = -40\)

Step 7 ：Solve for y: \(y = \frac{-40}{8} = -5\)

Step 8 ：Substitute \(y = -5\) into equation 4, we get: \(x = -3 - (-5) = 2\)

Step 9 ：So, the solution to the system of equations is \(x = 2\), \(y = -5\), \(z = 3\)

Step 10 ：\(\boxed{x = 2, y = -5, z = 3}\)